Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
Answer:
1.41 moles H2O2(with sig figs)
Explanation:
okay so what is the molar mass of H2O2= (1.008 g/mol)2+(16.00g/mol)2= (2.016+ 32.00) g/ mol
= 34. 02 g/mol
48.0g H2O2* 1 mol H2O2/ 34.02 g H2O2= 1.41 mol H2O2
Answer: B (to provide a statement that can be tested with an experiment
Answer: 
kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:
→ 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide: 
-1002.82 kJ/mol
Calcium chloride: -795.8 kJ/mol
Water: -285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)
So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
Answer:
It is the third one
Explanation:
"Both atoms achieve a more stable configuration."
